Using the Balanced Equation Stoichiometry Using the Balanced Equation

What does the balanced equation really mean? 2H2 + O2  2H2O Particles: Moles: 2 molecules H2 + 1 molecule O2  2 molecules H2O 2 moles H2 + 1 mole O2  2 moles H2O Which is more useful? moles

2H2 + O2  2H2O Like a recipe, the balanced equation tells you in what ratios the ingredients must be mixed. Therefore, for every 2 moles of H2 you will need 1 mole of O2. Mathematically, this can be written 2 moles H2 = 1 mole O2 What can you do with equalities? Dimensional Analysis

2H2 + O2  2H2O How many conversion factors (called mole ratios) can be written from this 1 equation? 2 moles H2 = 1 mole O2 2 moles H2 = 2 mole H2O 1 mole O2 = 2 moles H2O

How many moles of O2 are needed to react with 6.4 mol H2? 2H2 + O2  2H2O How many moles of O2 are needed to react with 6.4 mol H2? = 6.4 mol H2 1 mol O2 3.2 mol O2 2 mol H2

How many moles of H2O are produced from 6.4 mol H2? 2H2 + O2  2H2O How many moles of H2O are produced from 6.4 mol H2? = 6.4 mol H2 2 mol H2O 6.4 mol H2O 2 mol H2

Multi-step Stoichiometry EVERY stoichiometry problem involves moving from 1 substance to another! How many molecules of H2 does it take to make 647 g of H2O? 2H2 + O2  2H2O ? molecules 647 g The only way to move from 1 substance to another is the mole ratio!

Multi-step Stoichiometry You have three tools for stoichiometry problems! Tools of Stoichiometry 1. Mole Ratio balanced equation coefficients moles A  moles B 2. Molar Mass periodic table mass = 1 mol mass A (g)  1 mole A Avogadro's # 6.02  1023 particles = 1 mol particles particles A  1 mole A Volume 22.4 L at STP 1 mole A  22.4 L of A Every stoichiometry problem uses the mole ratio!!!

Multi-step Stoichiometry To answer the original question, you need to use all 3 tools. How many molecules of H2 does it take to make 647 g of H2O? 2H2 + O2  2H2O 647 g ? molecules This problem not only involves changing substances (mole ratio), but also mass (molar mass) and molecules (Avogadro’s number).

2H2 + O2  2H2O ? molecules 647 g First, you want to set up a dimensional analysis problem. Put your given and target in the equation. 647 g H2O = ? molecules H2

Next, convert g of H2O to moles of H2O using molar mass. 2H2 + O2  2H2O ? molecules 647 g Next, convert g of H2O to moles of H2O using molar mass. 647 g H2O = ? molecules H2 1 mol H2O 18.0153 g H2O

Next, EVERY problem must have a mole ratio step, so put that next! 2H2 + O2  2H2O ? molecules 647 g Next, EVERY problem must have a mole ratio step, so put that next! 647 g H2O = ? molecules H2 2 mol H2 1 mol H2O 2 mol H2O 18.0153 g H2O You don’t want H2O, so put moles H2O in the bottom of the middle step (remember you have to be in moles to move from 1 substance to another). You want to change to H2, so put moles of H2 on top (mole ratio is moles to moles). The numbers come from the balanced equation!

Answer = 2.161025 molecules H2 2H2 + O2  2H2O 647 g Since, we need molecules and the middle step gets us to moles H2, we need to use Avogadro’s number. 647 g H2O = ? molecules H2 mol H2O mol H2 2 g H2O 18.0153 1 6.021023 molecule H2 1 mol H2 Answer = 2.161025 molecules H2

Calculating Stoichiometric Problems Balance the equation Convert given to moles Set up mole ratios. Use mole ratios to calculate moles of desired chemical. Convert moles to target.

Another Example How many grams of O2 are needed to use up 13.6 moles of H2? 2H2 + O2  2H2O 13.6 mol g ? 31.9988 g O2 13.6 mol H2 1 mol O2 = g O2 218 2 mol H2 1 mol O2

12.3 Limiting Reactants Cake Recipe 2 cups flour 1 cup butter 1tsp. salt 1/2 cup sugar If I have 6 cups of flour, 4 cups of butter, 112 tsp. of salt, and 429 cups of sugar, how many cakes can I make?

12.3 Limiting Reactants Cake Recipe You Have in the Cupboard 2 cups flour 6 cups flour 1 cup butter 4 cup butter 1tsp. Salt 112 tsp. salt 1/2 cup sugar 429 cups sugar 6 cups flour 1 cakes = cakes 3 2 cups flour 4 cups butter 1 cakes = cakes 4 1 cups butter

12.3 Limiting Reactants Cake Recipe You Have in the Cupboard 2 cups flour 6 cups flour 1 cup butter 4 cup butter 1tsp. Salt 112 tsp. salt 1/2 cup sugar 429 cups sugar 112 tsp. salt 1 cakes = cakes 112 1 tsp. salt 1 cakes 429 cups sugar = cakes 858 .5 cups sugar

It also controls how much of the other ingredients we need! 12.3 Limiting Reactants Flour controls or limits how many cakes we can make because after we make 3 cakes we run out of flour. It also controls how much of the other ingredients we need!

12.3 Limiting Reactants How much butter is left over? First find how much was used and then subtract how much is left. FLOUR controls how much we use. 6 cups flour 1 cup butter = cups butter 3 2 cups flour 4 cups butter – 3 cups butter = 1 cup butter left

Chemical reactions work the same way!!!! 12.3 Limiting Reactants Limiting reactant - the reactant that makes the least. It controls everything about the reaction because when it is used up the reaction stops. Chemical reactions work the same way!!!!

12.3 Limiting Reactants 2H2 + O2  2H2O 2.50 g 3.40 g How much water can be made? 2.50 g H2 1 mol H2 2 mol H2O 18.0153 g H2O You must find which reactant controls (limiting reactant) the reaction. = g H2O 22.3 2.01588 g H2 2 mol H2 1 mol H2O 3.40 g O2 1 mol O2 2 mol H2O 18.0153 g H2O = g H2O 3.83 31.9988 1 mol O2 1 mol H2O g O2 O2 is therefore the limiting reactant and 3.83 g of water is produced!

12.3 Limiting Reactants 2H2 + O2  2H2O How much H2 is left over? 2.50 g 3.40 g How much H2 is left over? Find out how much H2 was used up by the 3.40 g O2 Then subtract that amount from 2.50 g to find out how much H2 is left

12.3 Limiting Reactants 2H2 + O2  2H2O How much H2 is left over? 2.50 g 3.40 g How much H2 is left over? 3.40 g O2 1 mol O2 2 mol H2 2.01588 g H2 = g H2 .428 31.9988 1 mol O2 1 mol H2 g O2 2.50 g .428 g - 2.07 g H2 left over

12.4 Percent Yield

We made theoretical prediction of how much could be made. Percent Yield When we calculated that 3.83 g of H2O could be made in the previous sample problem, we didn’t actually perform the reaction and measure the resulting water. We made theoretical prediction of how much could be made. Actual Yield – do the reaction to find the yield Theoretical Yield – do the math to predict the yield

Under ideal conditions, this is exactly what will happen. 12.4 Percent Yield When doing stoichiometric calculations, you are calculating what is known as a theoretical yield - the maximum amount that could be produced during the reaction. Under ideal conditions, this is exactly what will happen. In reality, however, the actual yield will be less than what is predicted.